Question

In an $AC$ circuit, voltage $V = V_0sin\omega t$ and inductor $L$ is connected across the circuit. Then the instantaneous power will be

• A. $\frac{V_{0}^{2}}{2\omega L}$sin $\omega t$
• B. $\frac{-V_{0}^{2}}{2\omega L}$sin $\omega t$
• C. $\frac{-V_{0}^{2}}{2\omega L}$sin $2\omega t$
• D. $\frac{V_{0}^{2}}{\omega L}$sin $2\omega t$

Answer 1

Answer: (C)

If $\frac{dI}{dt}$ is the rate of change of current through inductor L of any instant, then induced emf in the inductor of the same instant is
$\quad\quad=\left(-\frac{L dI}{dt}\right)$
To maintain the flow of current in the circuit applied voltage must be equal and opposite to the induced voltage.
$\therefore \quad V=-\left(-L \frac{dI}{dt}\right).$
Given, $V=V_{0} sin \omega t\quad\quad\quad\quad ... \left(i\right)$
$\therefore \quad V_{0} sin \omega t=\frac{L dI}{dt}$
$\quad\quad \frac{dI}{dt}=\frac{V_{0}}{L} sin \omega t$
or $I=\frac{-V_{0} }{L} \frac{cos \omega t}{\omega}\quad\quad\quad\quad\quad\quad ... \left(ii\right)$
Instantaneous power,
$∵ P=VI$
$\quad=\left(V_{0} sin \omega t\right)\left(-\frac{V_{0} cos \omega t}{L\omega}\right)\quad\left(Using \left(i\right) \&\left(ii\right)\right)$
$\quad=\frac{-V^{2}_{0} sin 2\omega t}{2L \omega}$